T3 Property is Hereditary

Theorem

Let $T = \struct {S, \tau}$ be a topological space which is a $T_3$ space.

Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.

Then $T_H$ is a $T_3$ space.

Proof

Let $T = \struct {S, \tau}$ be a $T_3$ space.

Then:

$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$

That is, for any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.

We have that the set $\tau_H$ is defined as:

$\tau_H := \set {U \cap H: U \in \tau}$

Let $F \subseteq H$ such that $F$ is closed in $H$.

Let $y \in H$ such that $y \notin F$.

From Closed Set in Topological Subspace $F$ is also closed in $T$.

Because $T$ is a $T_3$ space, we have that:

$\exists U, V \in \tau: F \subseteq U, y \in V, U \cap V = \O$

As $F \subseteq H$ and $y \in H, y \notin F$ we have that:

$F \subseteq U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$

and so the $T_3$ axiom is satisfied in $H$.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces